Let $p$ be a prime and let $G$ be a group of order $p^{a}m$ where $p$ does not divide $m$. Assume $P$ is a subgroup of $G$ of order $p^a$ and $N$ is a normal subgroup of $G$ of order $p^{b}n$ where $p$ does not divide $n$.
Prove that $\vert P\cap N\vert = p^b$ and $\vert PN/N\vert=p^{a-b}$
So I know that since $N$ is normal in $G$ and $P$ is a subgroup that $PN$ is a subgroup of $G$. And I know that $P\cap N\trianglelefteq N$ and $P\cap N\trianglelefteq P$.
So then $\vert P\cap N\vert$ divides the order of $P$
Thus $\vert P\cap N\vert=p^c$
Then I want to find $\vert N:P\cap N\vert$
Since $P\cap N$ is normal in $N$ its cosets partition $N$
I want to prove there must be $n$ cosets so that $c=b$
I know there must be at least $n$ but I don't see why there can't be $p^dn$ where $p^b=p^{c+d}$
After proving that then I can show the rest.
Since if $\vert N:P\cap N\vert=n$ then $\vert P\cap N\vert = p^b$ and I get that
$PN/N\cong P/P\cap N$ by second isomorphism theorem. which shows $\vert PN/N\vert = p^{a-b}$
From the Second Isomorphism Theorem: $$|PN| = \frac{|P||N|}{|P \cap N|} $$ You can observe that $P$ and $N$ are subgroups of $PN$. Thus, by Lagrange's Theorem, $p^an \mid |PN|$. We can also observe that from above, $|PN|$ divides $|P||N|=p^{a+b}n$. Finally, by Lagrange's Theorem once again, we have $|PN|$ divides $|G|=p^am$. These three conclude that $|PN|=p^an$. Substitution yields $|P \cap N|=p^b$.